People doing math. :eek:
In their spare time. :eek: :eek:
For fun. :eek: :eek: :eek:
[Piper slowly backs out of the room, not making any sudden moves.]
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Piper, I’ve been known to convert numbers into base 8 in my head just because I have nothing else to do, and in my spare moments I occasionally try to visualize five- and six-dimensional objects.
And everyone else in this thread beat me to the answer.
Be very afraid.
[/hijack]
I, Brian: perhaps you’re thinking of 1-(1/3)+(1/5)-(1/7)+(1/9)-…, which converges to pi/4.
Here is another version of Karl Gauss’ proof from Mathematics for Scientists and Engineers, Sokolnikof and Redheffer.
The series 1/n as n goes from 1 to n can be arranged as follows:
1+1/2+(1/3+1/4)+(1/5+1/6+1/7+1/8)+(1/9+…+1/16)+…
compare this term by term with:
1+1/2+(1/4+1/4)+(1/8+1/8+1/8+1/8)+(1/16+…+1/16)+…
Every term in the top sequence is equal to or greater than the corresponding term in the bottom sequence.
The bottom sequence can also be written as:
1+1/2+1/2+1/2+1/2+… which obviously increases without limit, i.e. diverges. The first sequence must diverge also since all the terms in it are at least as big as the terms in this sequence,
You have 3000+ posts, and you’re shocked to see a math thread?
Anyway, can anyone confirm or debunk my method (integrals)? The fact that I thought of it makes me think its wrong, so I’m really curious as to how smart I should feel.
In this case, your method is pretty close. But if f varies a lot between integers, the integral and the sum can be pretty far off. Consider f(x) = x + |40,000 * sin(2[symbol]p[/symbol]x)|. The sum of f(x) over [1, 10] is 55. The integral of f(x) over the same interval is (50[symbol]p[/symbol] + 800000)/[symbol]p[/symbol], which is roughly 255000.
I should add that if you were to accidentally mistype f without the absolute value and take the integral of that, you’d get 50. It’s cause there’s a lot of area under the curve in that case.
(With apologies to Northern Piper.)
Qwertyasdfg, here’s how to do summation approximation using integrals. First the function must by strictly increasing or decreasing. If not, then you have to work on it piecewise, assuming that is practical, and no asymptotes in the immediate area. 1/n fits the criteria if we are careful to stay away from n=0.
Think of building a bar chart using 1/n. Make a bar, width 1, of height 1/1 between 1 and 2. Then another of height 1/2 between 2 and 3, then height 1/3, etc. Finish with one height 1/n between n and n+1.
Note that the sum is equal to the areas of these bars. Plot the curve 1/n over this and note that the area under the 1/n curve is a little less than the sum. So a lower bound for the sum is the integral from 1 to n+1 (!) of 1/n. (Yeah, I shouldn’t use n twice like this.)
For an upper bound, plot 1/(n-1) instead. Oops, n=1 looks bad. But the rest of the sum, from 2 to n, is less than the integral from 2 to n+1 of 1/(n-1). Add in a +1 for the missing part.
If you do the math right, you’ll see that the upper and lower bound are ln n + lower order stuff. (You gotta know how to approximate ln (n-1). That’s another “lecture.”)
So the derivative of any nicely behaved, slow growing function, can be plugged into a summation that diverges. E.g., the derivative of ln ln ln ln n in a sum will also diverge, and much more slowly too.
(“More slowly”, “slower”? Ah forget it, I’m going to watch “The Simpsons.”)